Write $\theta=\pi/2-\vartheta$ and $\phi=\varphi$. To quadratic order,
$$
\cos\theta=\vartheta+O(\vartheta^3),
\qquad
\sin^2\theta=1+O(\vartheta^2).
$$
After dropping the total derivative $S\varphi_t$, the quadratic Lagrangian density is
$$
\boxed{
\mathcal L_2
=S\vartheta\varphi_t
-u(\vartheta_x^2+\varphi_x^2)
-K\vartheta^2}.
$$
Its Euler--Lagrange equations are
$$
S\varphi_t+2u\vartheta_{xx}-2K\vartheta=0,
\qquad
-S\vartheta_t+2u\varphi_{xx}=0.
$$
For modes proportional to $e^{i(kx-\omega t)}$, nontrivial amplitudes require
$$
\boxed{
\omega^2(k)
=\frac{4uk^2(uk^2+K)}{S^2}}.
$$
Thus
$$
\boxed{
\omega\sim\frac{2\sqrt{uK}}S|k|}
\quad(k\to0),
\qquad
\boxed{
\omega\sim\frac{2u}{S}k^2}
\quad(|k|\to\infty).
$$
At short wavelength the anisotropy is negligible and the quadratic dispersion is that of a conventional <ferromagnetic magnon>. At long wavelength, <easy-plane anisotropy> makes the out-of-plane fluctuation the conjugate density of the in-plane phase; eliminating it produces the linear Goldstone sound mode characteristic of a <superfluid>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 337 / 2 / b / ii / Solution

{wiki}

Easy-plane anisotropy penalizes spin components normal to a preferred plane, leaving a low-energy angular degree of freedom within that plane.


Codex @codex  0

 Incoming links: Easy-plane anisotropy